An Area Worth Exploring
It's common knowledge that the graph of
x
2
+
y
2
≤
1
is a unit disk, and the area of the region is
π
.
However, it's not common for someone to know the shape of the graph ⌊ x 2 ⌋ + ⌊ y 2 ⌋ = 1 , where ⌊ ⋅ ⌋ is the floor function . To three decimal places, what is the area of the region on the coordinate plane that satisfies the equation ⌊ x 2 ⌋ + ⌊ y 2 ⌋ = 1 ?
Note:
Try to draw a hand sketch of this curve and provide analysis instead of using any software!
The answer is 3.3137.
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3 solutions
Could you explain where the value 2 came from?
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any value near and lower than 2 [i.e: 2 − 0 . 0 0 1 ] will, when passed to the squared function, be evaluated to a value near but lower than 2 (i.e. 1.999). The floor function will then round this down to 1. 2 is the maximum value (but not 2 itself) that x or y can be so that, after it is squared AND the floor function is applied, the number you get is 1 :-)
I followed the same logic and put in the the answer 3.31, which it rejected, then tried 3.314 when that didn't work. What do they mean by "three decimal places" here?
In the question should be =2 , not =1
Feel cheated when it says 7.3137085 is not the correct answer
Relevant wiki: Floor and Ceiling Functions - Problem Solving
x 2 ≥ 0 , ⌊ x 2 ⌋ = 0 or 1 . As ⌊ x 2 ⌋ + ⌊ y 2 ⌋ = 1 , we can conclude the solution is either ⌊ x 2 ⌋ = 1 , ⌊ y 2 ⌋ = 0 , or ⌊ x 2 ⌋ = 0 , ⌊ y 2 ⌋ = 1 .
If ⌊ x 2 ⌋ = 0 , then 0 ≤ x 2 < 1 , so − 1 < x < 1 . If ⌊ x 2 ⌋ = 1 , then 1 ≤ x 2 < 2 , so − 2 < x ≤ − 1 or 1 ≤ x < 2 . Same for y .
Now, we can conclude that the diagram is the combine of − 1 < x < 1 , − 1 < x < 1 , 2 < x ≤ − 1 , 1 ≤ x < 2 , − 2 < y ≤ − 1 1 ≤ y < 2 − 1 < y < 1 − 1 < y < 1 Draw the diagram, we find that it is combine of 4 non-overlap rectangle with same area.
Hence, the answer is 4 ( 2 − 1 ) ( 2 ) = 3 . 3 1 3 7 .
The question was asking what was the region bounded by the curve, not the area in the curve. So shouldn't the answer also include the 4-unit square inside (bounded by) those 4 rectangles and thus be 7.3137?
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Oh yah, I agree.
The coordinate of the points in these four squares don't fit that equation. So they can't be calculate inside.
You threw me off there for a while when you "omitted as an exercise to the reader" a minus before the first x in the first rectangular region.
Relevant wiki: Floor and Ceiling Functions - Problem Solving
Since ⌊ x 2 ⌋ and ⌊ y 2 ⌋ can take only zero and positive integer values, only possible pairs to fit the equation are ( ⌊ x 2 ⌋ , ⌊ y 2 ⌋ ) = { ( 1 , 0 ) ( 0 , 1 ) } .
If ⌊ x 2 ⌋ = 0 , then 0 ≤ x 2 < 1 , then x ∈ ( − 1 , 1 )
then ⌊ y 2 ⌋ = 1
⟹ 1 ≤ y 2 < 2 ⟹ y ∈ ( − 2 , − 1 ] ∪ [ 1 , 2 ) .
Same thing holds for 0 ≤ y < 1 .
When we graph it, we get four rectangles with sides 2 and 2 − 1 , thus the area of the region on the coordinate plane which satisfies the equation is 4 × 2 × ( 2 − 1 ) ≈ 3 . 3 1 4
That is a strange figure indeed. Nice solution!
This is what the graph approximately looks like. When y = ( − 1 , 1 ) => x = [ 1 , 2 ) and ( − 2 , − 1 ] . Vice versa for x = ( − 1 , 1 ) . Each of the four rectangles have h e i g h t = 2 and w i d t h = 2 − 1 . Since there are four, a r e a = 4 ∗ 2 ∗ ( 2 − 1 ) which is approximately 3 . 3 1 3 7 0 8 4 9 8 9 8